2004 AMC 12B Problem 24

Let $\angle DBE = \alpha$ and $\angle DBC = \beta.$ Since $\overline{BD}$ is the altitude of the isosceles triangle, $\angle CBE = \alpha - \beta$ and $\angle ABE = \alpha + \beta.$ The geometric progression gives $\tan(\alpha - \beta)\tan(\alpha + \beta) = \tan^2\alpha,$ which simplifies to $\tan^2\beta(\tan^4\alpha - 1) = 0,$ so $\tan\alpha = 1$ and $\alpha = 45^\circ.$ Writing $DC = a$ and $BD = b,$ the arithmetic progression $\cot\angle DBE,$ $\cot\angle CBE,$ $\cot\angle DBC$ becomes $1, \dfrac{b + a}{b - a}, \dfrac{b}{a},$ forcing $b = 3a.$ With $BE = 10$ and $\angle DBE = 45^\circ,$ we get $b = \dfrac{BE}{\sqrt2} = 5\sqrt2,$ so $a = \dfrac{5\sqrt2}{3}.$ The area of $\triangle ABC$ is $\tfrac12 (AC)(BD) = ab$ $= 5\sqrt2 \cdot \dfrac{5\sqrt2}{3}$ $= \dfrac{50}{3}.$ Thus, the correct answer is B.